How do you find the equations for the normal line to y=x^2y=x2 through (2,4)?

1 Answer
Steve M
Dec 13, 2016

y=-1/4x+9/2y=−14x+92

Explanation:

The gradient of the tangent to a curve at any particular point is give by the derivative of the curve at that point. The normal is perpendicular to the tangent, so the product of their gradients is -1−1

so If y=x^2y=x2 then differentiating wrt xx gives us:

dy/dx = 2xdydx=2x

When x=2 => y=2^2=4x=2⇒y=22=4 (so (2,4)(2,4) lies on the curve)
and dy/dx=(2)2=4dydx=(2)2=4

So the normal we seek passes through (2,4)(2,4) ad has gradient -1/4−14 so using y-y_1=m(x-x_1)y−y1=m(x−x1) the equation we seek is;

y-4=-1/4(x-2) y−4=−14(x−2)
:. y-4=-1/4x+1/2
:. y=-1/4x+9/2

We can confirm this graphically:
enter image source here

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